University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 28: 50



Work Step by Step

$$\sin^2\frac{3\pi}{8}$$ *Recall the half-angle formula for sine, which is $$\sin^2\theta=\frac{1-\cos2\theta}{2}$$ Thus, $$\sin^2\frac{3\pi}{8}=\frac{1-\cos\frac{3\pi}{4}}{2}$$ Angle $\frac{3\pi}{4}$ is the equivalent of angle $\frac{\pi}{4}$ but in the second quadrant in the unit circle, where cosine is negative. Therefore, $\cos\frac{3\pi}{4}=-\cos\frac{\pi}{4}=-\frac{\sqrt2}{2}$ So, continuing with the given function: $$\sin^2\frac{3\pi}{8}=\frac{1-\Big(-\frac{\sqrt2}{2}\Big)}{2}$$ $$\sin^2\frac{3\pi}{8}=\frac{1+\frac{\sqrt2}{2}}{2}$$ $$\sin^2\frac{3\pi}{8}=\frac{\frac{2+\sqrt2}{2}}{2}$$ $$\sin^2\frac{3\pi}{8}=\frac{2+\sqrt2}{4}$$
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